Jul 10, 2012 - CA] 10 Jul 2012 ...... It is a nice fact (see [8] or [4, p.159]) that the maximal operator associated with ..... Mathématique de France, Paris (1995).
arXiv:1207.2407v1 [math.CA] 10 Jul 2012
The maximal singular integral: estimates in terms of the singular integral Joan Verdera Abstract This paper considers estimates of the maximal singular integral T ∗ f in terms of the singular integral T f only. The most basic instance of the estimates we look for is the L2 (Rn ) inequality kT ∗ f k2 ≤ C kT f k2 . We present the complete characterization, recently obtained by Mateu, Orobitg, P´erez and the author, of the smooth homogeneous convolution Calder´on–Zygmund operators for which such inequality holds. We focus attention on special cases of the general statement to convey the main ideas of the proofs in a transparent way, as free as possible of the technical complications inherent to the general case. Particular attention is devoted to higher Riesz transforms.
1 Introduction In this expository paper we consider the problem of estimating the Maximal Singular Integral T ∗ f only in terms of the Singular Integral T f. In other words, the function f should appear in the estimates only through T f. The context is that of classical Calder´on– Zygmund theory: we deal with smooth homogeneous convolution singular integral operators of the type Z T f (x) = p.v. f (x − y) K(y) dy ≡ lim T ǫ f (x) , (1) ǫ→0
where ǫ
T f (x) =
Z
f (x − y)K(y) dy
|y−x|>ǫ
is the truncated integral at level ǫ. The kernel K is Ω(x) , x ∈ Rn \ {0} , (2) |x|n where Ω is a (real valued) homogeneous function of degree 0 whose restriction to the unit sphere S n−1 is of class C ∞ (S n−1 ) and satisfies the cancellation property Z Ω(x) dσ(x) = 0 , (3) K(x) =
|x|=1
1
σ being the normalized surface measure on S n−1 . The maximal singular integral is T ⋆ f (x) = sup |T ǫ f (x)|,
x ∈ Rn .
ǫ>0
As we said before, the problem we are envisaging consists in estimating T ⋆ f in terms of T f only. The well known Cotlar’s inequality T ⋆ f (x) 6 C (M(T f )(x) + M f (x)) ,
x ∈ Rn ,
(4)
is of no use because it contains the term f besides T f . The most basic form of the estimate we are looking for is the L2 inequality kT ⋆ f k2 6 CkT f k2 ,
f ∈ L2 (Rn ) .
(5)
This problem arose when the author was working at the David–Semmes problem ([2, p.139, first paragraph]). It was soon discovered ([7]) that the parity of the kernel plays an essential role. Some years after, a complete characterization of the even operators for which (5) holds was presented in [5] and afterwards the case of odd kernels was solved in [6]. Unfortunately there does not seem to be a way of adapting the techniques of those papers to the Ahlfors regular context in which the David–Semmes problem was formulated. The proof of the main result in [5] and [6] is long and technically involved. It is the purpose of this paper to describe the main steps of the argument in the most transparent way possible. We give complete proofs of particular instances of the main results of the papers mentioned, so that the reader may grasp, in a simple situation, the idea behind the proof of the general cases. Thus, in a sense, the present paper could serve as an introduction to [5] and [6]. Notice that (5) is true whenever T is a continuous isomorphism of L2 (Rn ) onto itself. Indeed a classical estimate, which follows from Cotlar’s inequality, states that kT ⋆ f k2 6 C k f k2,
f ∈ L2 (Rn ) ,
(6)
which combined with the assumption that T is an isomorphism gives (5). Thus (5) is true for the Hilbert Transform and for the Beurling Transform. The first non-trivial case is a scalar Riesz transform in dimension 2 or higher. Recall that the j-th Riesz transform is the Calder´on–Zygmund operator with kernel xj , |x|n+1
x ∈ Rn \ {0},
1 6 j 6 n.
The first non trivial case for even operators is any second order Riesz transform. For example, the second order Riesz transform with kernel x1 x2 , |x|n+2
x ∈ Rn \ {0}.
2
In Section 2 we prove the L2 estimate (5) for the second order Riesz transform above and in Section 4 for the j − th Riesz transform. Indeed, in both cases we prove a stronger pointwise estimate which works for all higher Riesz transforms. Recall that a higher Riesz transform is a smooth homogeneous convolution singular integral operator with kernel of the type P(x) , x ∈ Rn \ {0}, |x|n+d where P is a harmonic homogeneous polynomial of degree d > 1. The mean value property of harmonic functions combined with homogeneity yields the cancellation property (3). One has the following ([5]) Theorem 1. If T is an even higher Riesz transform, then x ∈ Rn ,
T ∗ f (x) 6 C M(T f )(x),
f ∈ L2 (Rn ) ,
(7)
where M is the maximal Hardy-Littlewood operator. Indeed, for a second order Riesz transform S one has that the truncation at level ǫ is a mean of S ( f ) on a ball. More precisely one has Z 1 ǫ S ( f )(y) dy (8) S ( f )(x) = |B(x, ǫ)| B(x, ǫ) A weighted variant of the preceding identity works for a general even higher Riesz transform. Of course, (5) for even higher Riesz transforms follows immediately from (7). It turns out that, as we explain in Section 3, (7) does not hold for odd Riesz transforms, not even for the Hilbert transform. But we can prove the following substitute result ([6]), which obviously takes care of (5) for odd higher Riesz transforms. Theorem 2. If T is an odd higher Riesz transform, then T ∗ f (x) 6 C M 2 (T f )(x),
x ∈ Rn ,
f ∈ L2 (Rn ) ,
(9)
where M 2 = M ◦ M is the iteration of the maximal Hardy- Littlewood operator. Without any harmonicity assumption the L2 estimate (5) does not hold. The simplest example involves the Beurling transform B, which is the singular integral operator in the plane with complex valued kernel −
11 1 z2 1 x2 − y2 1 2xy = − = − +i . 2 4 4 πz π |z| π |z| π |z|4
The Fourier transform of the tempered distribution p.v.(− π1 z12 ) is the function ξξ , so that B is an isometry of L2 (R2 ) onto itself. It turns out that the singular integral T = B + B2 = B(I + B) 3
does not satisfy the L2 control (5). The reason for that, as we will see later on in this Section, is that the operator I + B is not invertible in L2 (R2 ). One way to explain the difference between the even and odd cases is as follows. Theorem 1 concerns an even higher Riesz transform determined by a harmonic homogeneous polynomial of degree, say, d. In its proof one is lead to consider the operator (−△)d/2 , which is a differential operator. Instead, in Theorem 2, d is odd and thus (−△)d/2 is only a pseudo-differential operator. The effect of this is that in the odd case certain functions are not compactly supported and are not bounded. Nevertheless, they still satisfy a BMO condition, which is the key fact in obtaining the second iteration of the maximal operator. The search for a description of those singular integrals T of a given parity for which (5) holds begun just after [7] was published. The final answer was given in [5] and [6]. To state the result denote by A the Calder´on–Zygmund algebra consisting of the operators of the form λI + T , where T is a smooth homogeneous convolution singular integral operator and λ a real number. Theorem 3. Let T be an even smooth homogeneous convolution singular integral operator with kernel Ω(x)/|x|n . Then the following are equivalent. (i) T ∗ f (x) 6 C M(T f )(x),
x ∈ Rn ,
f ∈ L2 (Rn ),
where M is the Hardy-Littlewood maximal operator. (ii)
Z
∗
2
|T f | 6 C
Z
|T f |2 ,
f ∈ L2 (Rn ).
(iii) If the spherical harmonics expansion of Ω is Ω(x) = P2 (x) + P4 (x) + · · · ,
|x| = 1,
then there exist an even harmonic homogeneous polynomial P of degree d, such that P divides P2 j (in the ring of all polynomials in n variables with real coeficients) for all j, T = RP ◦ U, where RP is the higher Riesz transform with kernel P(x)/|x|n+d , and U is an invertible operator in the Calder´on–Zygmund algebra A. Several remarks are in order. First, it is surprising that the L2 control we are looking for, that is, condition (ii) above, is equivalent to the apparently much stronger pointwise inequality (i). We do not know any proof of this fact which does not go through the structural condition (iii). Second, condition (iii) on the spherical harmonics expansion of Ω is purely algebraic and easy to check in practice on the Fourier transform side. Observe that if condition (iii) is satisfied, then the polynomial P must be a scalar multiple of the first non-zero spherical harmonic P2 j in the expansion of Ω. We illustrate this with an example.
4
Example. Let P(x, y) = − π1 xy and denote by RP the second order Riesz transform in the plane associated with the harmonic homogeneous polynomial P. Its kernel is −
1 xy , π |z |4
z = x + iy ∈ C \ {0}.
(10)
According to a well known formula [9, p.73] the Fourier transform of the principal value distribution associated with this kernel is uv , ξ = u + iv ∈ C \ {0}. |ξ |2 This is also the symbol (or Fourier multiplier) of RP , in the sense that uv ˆ R[ f (ξ), ξ , 0, f ∈ L2 (Rn ). P ( f )(ξ) = |ξ |2 Similarly, the Fourier multiplier of the fourth order Riesz transform with kernel 2 x3 y − xy3 , π |z |6
z , 0,
is
u3 v − uv3 , ξ , 0. |ξ |4 Given a real number λ let T be the singular integral with kernel − Its symbol is
2 x3 y − xy3 1 2xy + λ . π |z |4 π |z |6 ! uv u2 − v2 1+λ . |ξ |2 |ξ|2
We clearly have T = RP ◦ U, 2
2
U being the bounded operator on L (R ) with symbol 1 + λ u |ξ|−v2 . Notice that the multiplier 2
2
n
2
1 + λ u |ξ|−v2 vanishes at some point of the unit sphere if and only if |λ| > 1. Therefore condition (iii) of Theorem 3 is satisfied if and only if |λ| < 1. For instance, taking λ = 1 one gets an operator for which neither the L2 estimate (ii) nor the pointwise inequality (i) hold. To grasp the subtlety of the division condition in (iii) it is instructive to consider the special case of the plane. The function Ω, which is real, has a Fourier series expansion iθ
Ω(e ) =
∞ X
inθ
cn e
=
n=−∞
=
∞ X
∞ X n=1
2 Re(cn einθ )
n=1
5
cn einθ + cn e−inθ
The expression 2 Re(cn ei nθ ) is the general form of the restriction to the unit circle of a harmonic homogeneous polynomial of degree n on the plane. There are exactly 2n zeroes of 2 Re(cn ei nθ ) on the circle, which are uniformly distributed. They are the 2n-th roots of unity if and only if cn is purely imaginary. Since Ω is even, only the Fourier coefficients with even index may be non-zero and so iθ
Ω(e ) =
∞ X
2 Re(c2 n ei 2 nθ ).
n=1
Replacing θ by θ + α we obtain i(θ+α)
Ω(e
)=
∞ X
2 Re(c2 n ei 2 nα ei 2 nθ ),
n=N
where c2 N , 0. Take α so that c2 N ei 2 N α is purely imaginary. Set γ2 n = c2 n ei 2 n α . Then i(θ+α)
Ω(e
)=
∞ X
2 Re(γ2 n ei 2 nθ ).
n=N
If Re(γ2 N ei 2 Nθ ) divides Re(γ2 n ei 2 nθ ), then , for some positive integer k, k
π π = , 4n 4N
or n = k N. This means that only the Fourier coefficients with index a multiple of 2 N may be non-zero : ∞ X Ω(ei(θ+α) ) = 2 Re(γ2 N p ei 2 N p θ ). p=1
Moreover γ2N p must be purely imaginary, that is, γ2 N p = r2 N p i, with r2 N p real. Replacing θ + α by θ we get iθ
Ω(e ) =
∞ X
2 Re(r2N p i e−i2N pα ei2N pθ ),
∞ X
r2N p i e−i2N pα ei2N pθ − r2N p i ei2N pα e−i2N pθ .
p=1
=
p=1
As it is well-known the sequence of the r2N p , p = 1, 2, . . . is rapidly decreasing, because Ω(eiθ ) is infinitely differentiable. Therefore the division property in condition (iii) of Theorem 1 can be reformulated as a statement about the arguments and the support of the Fourier coefficients of Ω(eiθ ). For odd operators the statement of Theorem 3 must be slightly modified ([6]).
6
Theorem 4. Let T be an odd smooth homogeneous convolution singular integral operator with kernel Ω(x)/|x|n . Then the following are equivalent. (i) T ∗ f (x) 6 C M 2 (T f )(x),
x ∈ Rn ,
f ∈ L2 (Rn ),
M 2 = M ◦ M being the iterated Hardy-Littlewood maximal operator. (ii)
Z
∗
2
|T f | 6 C
Z
|T f |2 ,
f ∈ L2 (Rn ).
(iii) If the spherical harmonics expansion of Ω is Ω(x) = P1 (x) + P3 (x) + · · · ,
|x| = 1,
then there exist an odd harmonic homogeneous polynomial P of degree d, such that P divides P2 j+1 (in the ring of all polynomials in n variables with real coeficients) for all j, T = RP ◦U, where RP is the higher Riesz transform with kernel P(x)/|x|n+d , and U is an invertible operator in the Calder´on–Zygmund algebra A. Sections 2 and 4 contain , respectively, the proofs of Theorems 1 and 2 for the most simple kernels. In Section 3 we show that the Hilbert transform does not satisfy the pointwise inequality (7). In Section 5 we prove that condition (iii) in Theorem 3 is necessary and in Section 6 that it is sufficient, in both cases in particularly simple situations. Section 7 contains brief comments on the proof of the general case and a mention of a couple of open problems.
2 Proof of Theorem 1 for second order Riesz transforms. For se sake of clarity we work only with the second order Riesz transform T with kernel x1 x2 . |x|n+2 The inequality to be proven, namely (7), is invariant by translations and by dilations, so that we only need to show that |T 1 f (0)| ≤ C M(T f )(0), where 1
T f (0) =
Z
Rn \B
(11)
x1 x2 f (x) dx |x|n+2
is the truncation at level 1 at the origin. Here B is the unit (closed) ball centered at the origin. A natural way to show (11) is to find a function b such that χRn \B (x)
x1 x2 = T (b). |x|n+2 7
One should keep in mind that T is injective but not onto. Then there is no reason whatsoever for such a b to exist. If such a b exists then Z Z 1 T f (0) = T b(x) f (x) dx = b(x) T ( f )(x) dx (12) If moreover b is in L∞ (Rn ) and is supported on B, we get Z 1 |T ( f )(x)| dx ≤ CM(T ( f ))(0). |T 1 f (0)| ≤ kbk∞ |B| |B| B Thus everything has been reduced to the following lemma. Lemma 5. There exists a bounded measurable function b supported on B such that χRn \B (x)
x1 x2 = T (b)(x), |x|n+2
for almost all
x ∈ Rn .
Proof. Let E be the standard fundamental solution of the Laplacian in Rn . Then, for some dimensional constant cn , we have that, in the distributions sense, ∂1 ∂2 E = cn p.v.
x1 x2 . |x|n+2
(13)
Let us define a function ϕ by E(x) ϕ(x) = A0 + A1 |x|2
on Rn \ B on B
(14)
where the constants A0 and A1 are chosen so that ϕ and ∇ϕ are continuous on Rn . This is possible because, for each i, x x ∈ Rn \ B cn |x|in , ∂i ϕ(x) = 2A1 xi , x∈B
and so, for an appropriate choice of A1 , the above two expressions coincide on ∂B for all i, or, equivalently, ∇ϕ is continuous. The continuity of ϕ is now just a matter of choosing A0 so that E(x) = A0 + A1 |x|2 on ∂B, which is possible because E is radial. The continuity of ϕ and ∇ϕ guaranties that we can compute a second order derivative of ϕ in the distributions sense by just computing it pointwise on B and on Rn \ B. The reason is that no boundary terms will appear when applying Green-Stokes to compute the action of the second order derivative of ϕ under consideration on a test function. Therefore ∆ϕ = 2nA1 χB ≡ b,
where the last identity is the definition of b. Since ϕ = E ∗ ∆ϕ we obtain, for some dimensional constant cn , ∂1 ∂2 ϕ = ∂1 ∂2 E ∗ ∆ϕ = cn p.v. 8
x1 x2 ∗ ∆ϕ = cn T (b). |x|n+2
On the other hand, by (14) and noticing that ∂1 ∂2 |x|2 = 0, we get ∂1 ∂2 ϕ = χRn \B (x)cn
x1 x2 , |x|n+2
and the proof of Lemma 5 is complete.
Notice that (12) together with the special form of the function b found in the proof of Lemma 5 yield the formula (8), namely, that a truncation at level ǫ at the point x of S ( f ), S being a second order Riesz transform, is the mean of S ( f ) on the ball B(x, ǫ) .
3 The pointwise control of T ∗ by M◦T fails for the Hilbert transform We show now that the inequality H ∗ f (x) ≤ C M(H f )(x) ,
x∈R
f ∈ L2 (R),
(15)
where H is the Hilbert transform, fails. Replacing f by H( f ) in (15) and recalling that H(H f ) = − f , f ∈ L2 (R) , we see that (15) is equivalent to H ∗ (H( f ))(x) ≤ C M( f )(x) ,
x ∈ R,
f ∈ L2 (R).
It turns out that the operator H ∗ ◦ H is not of weak type (1, 1). Let us prove that if f = χ(0,1) , then there are positive constants m and C such that whenever x > m, log x H ∗ (H f )(x) ≥ C (16) x This shows that H ∗ ◦ H is not of weak type (1, 1). Indeed, choosing m > e if necessary, we have sup λ |{x ∈ R : H ∗ (H f )(x) > λ}| ≥ sup λ |{x > m : λ>0
λ>0
= C sup λ |{x > m : λ>0
log x > C −1 λ}| x
log x > λ}| ≥ C sup λ (ϕ−1 (λ) − e), x λ>0
where ϕ is the decreasing function ϕ : (e, ∞) → (0, e−1 ), given by ϕ(x) = logx x . To conclude observe that the right hand side of the estimate is unbounded as λ → 0: lim λϕ−1 (λ) = lim ϕ(λ)λ = ∞. λ→0
λ→∞
To prove (16) we recall that for f = χ(0,1) H f (y) = log 9
|y| . |y − 1|
Let m > 1 big enough to be chosen later on. Take x > m. By definition of H ∗ Z 1 |y| ∗ H (H f )(x) ≥ log dy |y−x|>m+x y − x |y − 1|
and splitting the integral in the obvious way Z −m Z ∞ 1 −y y 1 log dy + log dy −y + 1 y−1 −∞ y − x 2x+m y − x Z ∞ Z ∞ 1 y+1 y 1 log dy + log dy = A(x) + B(x), = y y−1 2x+m y − x m x+y where both A(x) and B(x) are positive. Hence H ∗ (H f )(x) ≥ A(x). Since
1 1 log(1 + ) ≈ , y y
as y → ∞,
there is a constant m > 1 such that whenever y > m 1 1 log(1 + y ) 3 < . < 1 2 2 y
Hence, for this constant m we have ! Z ∞ Z ∞ 1 dy 1 1 1 y ∞ log x dy ≈ A(x) = log 1 + = log , ≈ y x x+y m x m x+y y m x+y which proves (16). Notice that the term B(x) is better behaved : Z ∞ Z ∞ 1 y 2 dy 1 B(x) ≤ log dy ≤ ≤ . y−1 x 2x+m y − x 2x+m y y
4 Proof of Theorem 2 for first order Riesz transforms In this Section we prove that R∗j ( f )(x) ≤ C M 2 (R j ( f )),
x ∈ Rn ,
(17)
where R j is the j-th Riesz transform, namely, the Calder´on–Zygmund operator with kernel xj , |x|n+1
x ∈ Rn \ {0},
10
1 6 j 6 n.
Recall that M 2 = M ◦ M and notice that for n = 1 we are dealing with the Hilbert transform. The inequality (17) for the Hilbert transform is, as far as we know, new. To have a glimpse at the difficulties we will encounter in proving (17) we start by discussing the case of the Hilbert transform. As in the even case we want to find a function b such that 1 χ R\(−1,1) (x) = H(b). x Since H(−H) = I 1 b(x) = −H( χ R\(−1,1) (y))(x) y Z 1 1 1 dy = π |y|>1 y − x y =
1 |1 + x| log . πx |1 − x|
We conclude that, unlike in the even case, the function b is unbounded and is not supported in the unit interval (−1, 1). On the positive side, we see that b is a function in BMO = BMO(R), the space of functions of bounded mean oscillation on te line. Since b decays at infinity as 1/x2 , b is integrable on the whole line. However, the minimal decreasing majorant of the absolute value of b is not integrable, owing to the poles at ±1. This prevents a pointwise estimate of H ∗ f by a constant times M(H f ). We can now proceed with the proof of (17) keeping in mind the kind of difficulties we will have to overcome. We start with the analog of Lemma 5. We denote by BMO the space of functions of bounded mean oscillation on Rn . Lemma 6. There exists a function b ∈ BMO such that χRn \B (x)
xj = R j (b)(x), |x|n+1
for almost all
x ∈ Rn ,
1 6 j 6 n.
(18)
Proof. For an appropriate constant cn the function E(x) = cn satisfies
1 , |x|n−1
b = 1, E(ξ) |ξ|
0 , x ∈ Rn
0 , ξ ∈ Rn .
Since the pseudo-differential operator (−∆)1/2 is defined on the Fourier transform side as [ 1/2 ψ(ξ) = |ξ|ψ(ξ), ˆ (−∆) 11
E may be understood as a fundamental solution of (−∆)1/2 . This will allow to structure our proof in complete analogy to that of Lemma 5 until new facts emerge. Consider the function ϕ that takes the value cn on B and E(x) on Rn \ B . We have that ϕ = E ∗ (−∆)1/2 ϕ and we define b as (−∆)1/2 ϕ. As it is well known xj ∂ j E = −(n − 1)cn p.v. n+1 , |x| in the distributions sense and, since ϕ is continuous on the boundary of B, xj ∂ j ϕ = −(n − 1)cn χRn \B (x) n+1 (19) |x| also in the distributions sense. Then −(n − 1)cn χRn \B (x)
xj = ∂ jϕ |x|n+1 = ∂ jE ∗ b = −(n − 1)cn p.v.
xj ∗ b, |x|n+1
which is (18). It remains to show that b ∈ BMO. Checking on the Fourier transform side we easily see that 1/2
b = (−∆)
ϕ = γn
n X
Rk (∂k ϕ),
(20)
k=1
for some dimensional constant γn . Since ∂k ϕ is a bounded function by (19) and Rk maps L∞ into BMO, b is in BMO and the proof is complete. Unfortunately b is not bounded and is not supported on Rn \B. Moreover one can check easily that b blows up at the boundary of B as the function log(1/|1 − |x||). This entails that the the minimal decreasing majorant of the absolute value of b is not integrable, as in the one dimensional case. We take up now the proof of (17). By translation and dilation invariance we only have to estimate the truncation of R j f at the point x = 0 and at level ǫ = 1. By Lemma 6 Z Z xj 1 R j f (0) = − χRn \B (x) n+1 f (x) dx = − R j b(x) f (x) dx |x| Z = b(x) R j f (x) dx . Let b2B denote the mean of b on the ball 2B. We split the last integral above into three pieces Z Z Z 1 R j f (0) = (b(x) − b2B ) R j f (x) dx + b2B R j f (x) dx + b(x) R j f (x) dx 2B 2B Rn \2B (21) = I1 + I2 + I3 . 12
Since b2B is a dimensional constant the term I2 can be immediately estimated by C M(R j f )(0). The term I3 can easily be estimated if we first prove that |b(x)| 6 C
1 , |x|n+1
|x| > 2 .
(22)
Indeed, the preceding decay inequality yields Z 1 |I3 | 6 C |R j f (x)| n+1 dx 6 C M(R j f )(0) . |x| Rn \2B To prove (22) express b by means of (20) X X xk xk b X Rk ⋆ χRn \B (x) n+1 = = Rk ⋆ χB (x) n+1 Rk ⋆ Rk − γn k=1 |x| |x| k=1 k=1 n X xk = γn′ δ0 − Rk (χB (x) n+1 ) , |x| k=1 n
n
n
where γn′ is a dimensional constant and δ0 the dirac delta at the origin. The preceding formula for b looks magical and one may even think that some terms make no sense. For instance, the term Rk ⋆Rk should not be thought as the action of the k-th Riesz transform of the distribution p.v. xk /|x|n+1 . It is more convenient to look at it on the Fourier transform k side, where you see immediately that it is γn′ δ0 . The term Rk ⋆χB (x) |x|xn+1 should be thought as a distribution, which acts on a test function as one would expect via principal values (see below). If |x| > 1 we have Z xk − yk yk xk dy Rk (χB (x) n+1 )(x) = lim ǫ→0 ǫ
